Statistics Definitions > Hypergeometric Distribution

The hypergeometric distribution is a probability distribution that’s very similar to the binomial distribution. In fact, the binomial distribution is a very good approximation of the hypergeometric distribution as long as you are sampling 5% or less of the population.

Therefore, in order to understand the hypergeometric distribution, you should be very familiar with the binomial distribution. Plus, you should be fairly comfortable with the combinations formula.

If you need a brush up, see:

## Hypergeometric Distribution Formula

The (somewhat formal) definition for the hypergeometric distribution, where X is a random variable, is:

Where:

K is the number of successes in the population

k is the number of observed successes

N is the population size

n is the number of draws

You *could* just plug your values into the formula, but a much easier way is just to think through the problem, using your knowledge of combinations.

## Hypergeometric Distribution Example 1

A deck of cards contains 20 cards: 6 red cards and 14 black cards. 5 cards are drawn randomly *without replacement*. What is the probability that exactly 4 red cards are drawn?

The probability of choosing exactly 4 red cards is:

P(4 red cards) = # samples with 4 red cards and 1 black card / # of possible 4 card samples

Using the combinations formula, the problem becomes:

In shorthand, the above formula can be written as:

(6C4*14C1)/20C5

where

6C4 means that out of 6 possible red cards, we are choosing 4.

14C1 means that out of a possible 14 black cards, we’re choosing 1.

Solution = (6C4*14C1)/20C5 = 15*14/15504 = 0.0135

The binomial distribution doesn’t apply here, because the cards are not replaced once they are drawn. In other words, the trials are not independent events. For example, for 1 red card, the probability is 6/20 on the first draw. If that card is red, the probability of choosing another red card falls to 5/19.

## Hypergeometric Distribution Example 2

A small voting district has 101 female voters and 95 male voters. A random sample of 10 voters is drawn. What is the probability exactly 7 of the voters will be female?

101C7*95C3/(196C10)= (17199613200*138415)/18257282924056176 = 0.130

Where:

101C7 is the number of ways of choosing 7 females from 101 and

95C3 is the number of ways of choosing 3 male voters from 900*

196C10 is the total voters (196) of which we are choosing 10

*That’s because if 7/10 voters are female, then 3/10 voters must be male.

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PLEASE check the calculation , it does not seem to be right!!!!!!!!!!!!!!!!!

(17199613200*138415)/18257282924056176 = 0.130

Hi, Jarvis. I double checked the calculations and I’m getting the same answer. What did you get? Could you post your calculations? Thanks.